Two capacitors, C1 = 17.0 µF and C2 = 44.0 µF, are connected in series, and a 3.
ID: 1973217 • Letter: T
Question
Two capacitors, C1 = 17.0 µF and C2 = 44.0 µF, are connected in series, and a 3.0 V battery is connected across them.(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.
equivalent capacitance ____?____ µF
total energy stored ___?___ J
(b) Find the energy stored in each individual capacitor.
energy stored in C1 ___?___ J
energy stored in C2 ___?___ J
Show that the sum of these two energies is the same as the energy found in part (a). Will this equality always be true, or does it depend on the number of capacitors and their capacitances?
____?_____
(c) If the same capacitors were connected in parallel, what potential difference would be required across them so that the combination stores the same energy as in part (a)?
_____?_____ V
Explanation / Answer
a) 1/C = 1/C1 + 1/C2 C = C1 * C2 /(C1 + C2) = (17e-6 * 44e-6)/(17e-6 + 44e-6) = 12.26e-6 = 12.26 uF U = 0.5*C*V^2 = 0.5 * 12.26uF * 3^2 = 55.2 uJ = 5.52 * 10^-5 J b) Q = CV = 12.26u * 3 = 36.78 uC V1 = Q/C1 = 36.78 u /17 u = 2.164 V U1 = 0.5 * C * V^2 = 0.5 *17u * 2.1646^2 = 3.983 * 10^-5J V2 = 3 - 2.164 = 0.836 V U2 = 0.5 * C * V^2 = 0.5 *44u * 0.836^2 = 1.538 * 10^-5 J c) C1 + C2 = 3.983 * 10^-5 + 1.538 * 10^-5 =5.52 * 10^-5 J The energy stored in the capacitator is of the same value as part A. d) C = C1 + C2 = 17 u + 44 u = 61 uF U = 0.5 * C * V^2 V^2 = (2U)/C = 2 * 5.52 * 10^-5 / 61 uF = 1.725 V = 1.313 V
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